Talk:Orthogonal group

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[edit] Over a finite field

I think it would be useful to discuss the orthogonal groups over a finite field here. If someone is capable of writing such a section it would be greatly appreciated. TooMuchMath 17:31, 15 April 2006 (UTC)

This is absolutely essential. It's a shame that there is no section here on it. My knowledge of these groups isn't good enough to do a section, however. We also need some words on O+ and O- etc. Triangle e 14:26, 9 December 2006 (UTC)
I've added the most minimal amount of finite field stuff. I'd be grateful if someone could please expand. Triangle e 14:40, 22 April 2007 (UTC)
Things are progressing with orthogonal groups over finite fields. I removed some incorrect order formulas, and am working on giving the orders correctly and legibly. Here is a first draft, but only for odd characteristic. I'll work on the characteristic two case next. Both may require some reorganizing of the finite field sections (so that spinor and dickson come early enough to define SO and Ω).

For an odd prime power q: The orthogonal groups may refer to several particular families of groups, Ω < SO < O, each with index 2 in the next. The group Ω is the derived subgroup both of SO and of O, and Ω is usually a perfect group. One may define SO as the kernel of the determinant map on O, and Ω as the kernel of the spinor norm map on SO. Additionally, one is often interested in the quotients by the center, PΩ ≤ PSO ≤ PO, especially since PΩ is usually a simple group. For an odd prime power q, the orders of the orthogonal groups are given by:

Orthogonal group orders in odd characteristic
Odd dimensions Even dimensions
Group Order Group Order Group Order Group Order
O(2k+1, q) qk² · (q2k−1) · N · 2 PO(2k+1, q) qk² · (q2k−1) · N Oε(2k, q) qk(k−1) · (qk−ε) · N · 2 POε(2k, q) qk(k−1) · (qk−ε) · N
SO(2k+1, q) qk² · (q2k−1) · N PSO(2k+1, q) qk² · (q2k−1) · N SOε(2k, q) qk(k−1) · (qk−ε) · N PSOε(2k, q) qk(k−1) · (qk−ε) · N · ½
Ω(2k+1, q) qk² · (q2k−1) · N · ½ PΩ(2k+1, q) qk² · (q2k−1) · N · ½ Ωε(2k, q) qk(k−1) · (qk−ε) · N · ½ PΩε(2k, q) qk(k−1) · (qk−ε) · N · 1/gcd(4,qk−ε)
where N=\prod_{i=1}^{k-1} (q^{2i}-1), q is an odd prime power, and k is a positive integer.
Note the constant, messy parts are split off to make it easy to compare the orders. Perhaps some of this is silly, since the center is so small, people should be able to work out things for themselves based on just one of them, but which one depends heavily on the area in which the person is working. For me, O is useless, and only PΩ and Ω are interesting. JackSchmidt (talk) 23:05, 24 November 2007 (UTC)
Here is the even case.

For q a power of two: The orthogonal groups may refer to either of two families of groups, Ω ≤ O. The group Ω is the kernel of the Dickson pseudodeterminant; note that both the spinor norm and determinant are trivial for finite fields of characteristic 2. The group Ω is the derived subgroup of O, and Ω is usually a simple group. When the dimension of the underlying vector space is odd, then in fact O(2k+1,q) = Sp(2k,q). Note that in characteristic 2, the orthogonal groups have trivial center, so there is no distinction between O and PO, Ω and PΩ. The orders of the orthogonal groups are given by:

Orthogonal group orders in even characteristic
Group Order
O(2k+1, q) = PO(2k+1, q) =
SO(2k+1, q) = PSO(2k+1, q) =
Ω(2k+1, q) = PΩ(2k+1, q) =
Sp(2k+1, q) = PSp(2k+1, q)		 qk² · (q2k−1) · N
Oε(2k, q) = POε(2k, q) qk(k−1) · (qk−ε) · N · 2
Ωε(2k, q) = PΩε(2k, q) qk(k−1) · (qk−ε) · N, Except for ε=+1, k=2, q=2
where N=\prod_{i=1}^{k-1} (q^{2i}-1), q is a power of 2, and k is a positive integer,
ε=+1 if the form is hyperbolic, ε=-1 otherwise.

Small even dimensions and all odd dimensions have additional names:

  • Oε(2,q) = Dih(2(q−ε))
  • Ωε(2,q) is cyclic of order q−ε
  • O+1(4,q) = PSL(2,q) wreath Sym(2)
  • Ω+1(4,q) = PSL(2,q) × PSL(2,q), for q>2, but Ω+1(4,2) is an exception
  • Ω-1(4,q) = PSL(2,q²)
  • Ω+1(6,q) = PSL(4,q)
  • Ω-1(6,q) = PSU(4,q)
  • Ω(2k+1,q) = O(2k+1,q) = PSp(2k,q)
I wasn't sure of the best way to handle the weirdo O+(4,2), since its derived subgroup is smaller than expected. At any rate, I'll copy these into the article in some sane way this week so it will be easier to tweak. I just wanted to put some drafts up first in case someone had major stylistic concerns. The formulas are very detailed and easy to get wrong (different sources use different notations), so it is probably best not to adjust too much without checking carefully. JackSchmidt (talk) 16:43, 26 November 2007 (UTC)

[edit] Incorrect

The definition of the orthogonal group given here is FALSE. The orthogonal group is defined only over the real numbers. When defined over the complex numbers it is known as the unitary group. Over quaternions it is known as the spin group.

Let E the the n \times n identity matrix and

\mathbb{H} = \{a + ib + jc + kd : (a,b,c,d) \in \mathbb{R}^4, \ i^2 = j^2 = k^2 = -1, \ ij = k, \ jk = i, \ ki = j \} \ .

Then the CORRECT definitions are given as

O(n) = \{ X \in \mbox{Mat}(n,\mathbb{R}) : X^{\top}X = E \} \ ,
U(n) = \{ X \in \mbox{Mat}(n,\mathbb{C}) : \overline{X}^{\top}X = E \} \ ,
Sp(n) = \{ X \in \mbox{Mat}(n,\mathbb{H}) : \overline{X}^{\top}X = E \} \ .


The definition of the orthogonal group comes from finding all matrices which preserve the standard euclidean dot product, so the preserve distances and angles. The definition of the unitary group comes from finding matrices which preserve the standard Hermitian product. We see that

O(n) \subset U(n) \subset Sp(n) \ .

In particicular, the statement that \det(X) = \pm 1 is also INCORRECT. It is true for the othogonal group since the definition X^{\top}X = E means that \det(X^{\top}X) = \det(E) which in turn shows that det(X)2 = 1.

For the unitary group and the spin group we see that \det(\overline{X}^{\top}X) = \det(E) this means that

\overline{\det(X)}\det(X) = 1 \ , which means that | det(X) | = 1.

So for the unitary group, we see that

\det(X) \in \{ z \in \mathbb{C} : |z| = 1 \} \neq \{\pm 1\} \ .

Since \mathbb{H} is not commutative, the idea of a determinant is not well defined!

—The preceding unsigned comment was added by 128.101.10.93 (talk • contribs) .

You are confusing the complex orthogonal group (preserving a symmetric form) with the unitary group (preserving a hermitean form). R.e.b. 18:52, 5 May 2006 (UTC)

[edit] Incorrect order formulas

Above I posted a suggestion on presenting the order formulas, but my edit was reverted. Just to make sure everyone is on the same page, here are the wrong order formulas:

  • |O(2n+1,q)|=2q^n\prod_{i=0}^{n-1}(q^{2n}-q^{2i})
  • If − 1 is a square in \mathbf{F}_q, |O(2n,q)|=2(q^n-1)\prod_{i=0}^{n-1}(q^{2n}-q^{2i})
  • If − 1 is a nonsquare in \mathbf{F}_q, |O(2n,q)|=2(q^n+(-1)^{n+1})\prod_{i=0}^{n-1}(q^{2n}-q^{2i})

When n=1,q=3, the set of two by two matrices over the field with three elements whose inverse is their transpose has cardinality 8 as can be seen by writing down the 48 such invertible matrices and checking inverses. The first formula clearly doesn't apply, the second does not either but should give the order as 32, not a divisor of 48. The third formula is the one that should apply, but it gives the order as 64, definitely not a divisor of 48.

At any rate, correct order formula for all orthogonal groups are given above. In particular, O+(2,3) has order 4 and O-(2,3) has order 8. A reference is Grove's Classical Groups and Geometric Algebra, but I was planning on comparing references before including it in the article. JackSchmidt (talk) 21:55, 29 November 2007 (UTC)

Ups, you are right. The zero in the last two formulas should be a 1. —Preceding unsigned comment added by Franklin.vp (talk • contribs) 22:31, 29 November 2007 (UTC)

Thanks for checking the talk page. I checked the paper and it looks like this time the typo was not in the paper, but the published literature is filled with typos like this. That's why I said above to be careful checking the formulas; make sure to work some examples.
At some point, I will probably still want to replace the current article's order formulas with something like the tables above. However, your contribution is very sane and concrete and has a sane and concrete reference. I like {A:AA^t=I} much better than the Clifford algebras approach in Grove. It would be good to decide clearly when {A:AA^t=I} is O+ and when it is O-, hopefully with a reference (one can basically tell from the order formula). JackSchmidt (talk) 22:53, 29 November 2007 (UTC)